# If you decreased the volume of a sample of gas by a factor of three while maintaining a constant pressure, how would the absolute temperature of the gas be affected?

Jul 4, 2016

The temperature of the gas will decrease by a factor of $3$.

#### Explanation:

The pressure and temperature of a gas have a direct relationship when pressure and number of moles of gas are kept constant $\to$ this is known as Charles' Law.

Simply put, when the pressure of a given sample of gas is kept constant, increasing the volume of the gas will result in an increase in temperature

Similarly, decreasing the volume of the gas, as you have in your example, will result in a decrease in temperature.

Mind you, the temperature of the gas is actually its absolute temperature, i.e. its temperature expressed in Kelvin, $\text{K}$. Now, the thing to remember here is that decreasing or increasing the volume of the gas by a factor $x$ will cause the temperature to decrease or increase, respectively, by the same factor $x$.

Mathematically, this can be shown by using the equation for Charles' Law

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

${V}_{1}$, ${T}_{1}$ - the volume and temperature of the gas at an initial state
${V}_{2}$, ${T}_{2}$ - the volume and temperature of the gas at a final stat

Let's say that your starting volume is ${V}_{1}$. Decreasing the volume by a factor of $3$ means getting it down to

${V}_{2} = \frac{1}{3} \cdot {V}_{1} \to$ the volume decreases by a factor of $3$

Rearrange the equation to solve for ${T}_{2}$, the second temperature of the gas.

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \implies {T}_{2} = {V}_{2} / {V}_{1} \cdot {T}_{1}$

You will thus have

${T}_{2} = \frac{\frac{1}{3} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}} \cdot {T}_{1} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{1}{3} \cdot {T}_{1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As you can see, decreasing the volume by a factor of $3$ will cause the absolute temperature to decrease by a factor of $3$.